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1 answer:
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Explanation:
Gravity is the force of attraction between two objects. It depends upon the mass of the objects and the distance between the objects. Mathematically, the force of gravity is given by :
Where
G is the universal gravitational constant
are masses
d is the distance between two masses
So, statement (2) describes gravity "Gravity is the force of attraction between two objects; it is dependent upon the mass of the objects and the distance between the objects".
Answer:
<em>radius of the loop = 7.9 mm</em>
<em>number of turns N ≅ 399 turns</em>
Explanation:
length of wire L= 2 m
field strength B = 3 mT = 0.003 T
current I = 12 A
recall that field strength B = μnI
where n is the turn per unit length
vacuum permeability μ = = 1.256 x 10^-6 T-m/A
imputing values, we have
0.003 = 1.256 x 10^−6 x n x 12
0.003 = 1.507 x 10^-5 x n
n = 199.07 turns per unit length
for a length of 2 m,
number of loop N = 2 x 199.07 = 398.14 ≅ <em>399 turns</em>
since there are approximately 399 turns formed by the 2 m length of wire, it means that each loop is formed by 2/399 = 0.005 m of the wire.
this length is also equal to the circumference of each loop
the circumference of each loop =
0.005 = 2 x 3.142 x r
r = 0.005/6.284 = = 0.0079 m =<em> 7.9 mm</em>
Answer:
The magnitude of the final velocity is approximately 0.526 m/s in approximately the direction of 8.746° East of South
Explanation:
The given collision parameters are;
The kind of collision experienced by the four velcro-lined air-hockey disk = Inelastic collision
The mass of the first disk, m₁ = 50.0 g
The velocity of the first disk, v₁ = 0.80 m/s West = -0.8·i
The mass of the second disk, m₂ = 60.0 g
The velocity of the second disk, v₂ = 2.50 m/s North = 2.5·j
The mass of the third disk, m₃ = 100.0 g
The velocity of the third disk, v₃ = 0.20 m/s East = 0.20·i
The mass of the fourth disk, m₄ = 40.0 g
The velocity of the fourth disk, v₄ = 0.50 m/s South = -0.50·j
Therefore, the total initial momentum of the four velcro-lined air-hockey disk, is given as follows;
= m₁·v₁ + m₂·v₂ + m₃·v₃ + m₄·v₄ = 50.0×(-0.80·i) + 60.0×(2.50·j) + 100 × (0.20·i) + 40.0 × (-0.50·j)
∴ = -40·i + 150·j + 20·i - 20·j = -20·i + 130·j
∴ = -20·i + 130·j
By the law of conservation of linear momentum, we have;
= -20·i + 130·j
Therefore, given that the collision is perfectly inelastic, the disks move as one after the collision and the four masses are added to form one mass, "m", m = m₁ + m₂ + m₃ + m₄ = 50.0 + 60.0 + 100.0 + 40.0 = 250.0
∴ m = 250.0 g
Let, "v" represent the final velocity of the four disks moving as one after the collision
We have;
= m × v = 250.0 × v = -20·i + 130·j
∴ v = -20·i/250 + 130·j/250 = -0.08·i + 0.52·j
The final velocity of the four disks after collision, v = -0.08·i + 0.52·j
The magnitude of the final velocity, = √((-0.08)² + (0.52)²) ≈ 0.526
≈ 0.526 m/s
The direction of the final velocity, θ = arctan(0.52/(-0.08)) ≈ -81.254°
The direction of the final velocity, θ ≈ -81.254° which is 8.746° East of South